Question

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is given in the following problems. State if the ratio test is inconclusive. $$ \sum_{n=1}^{\infty} \frac{(2 n) !}{(n / e)^{2 n}} $$

Short answer

The series diverges by the Ratio Test.

Step-by-step solution

Identify the General Term

First, identify the general term of the series: \(a_n = \frac{(2n)!}{(n/e)^{2n}}\). This is the term by term expression for the series given in the problem.

Apply the Ratio Test Expression

The ratio test involves calculating the limit of the absolute value of \(\frac{a_{n+1}}{a_n}\) as \(n\rightarrow \infty\). First, find \(a_{n+1}\):\[a_{n+1} = \frac{(2(n+1))!}{((n+1)/e)^{2(n+1)}}.\]

Compute \(\frac{a_{n+1}}{a_n}\)

Substitute the expressions for \(a_{n+1}\) and \(a_n\) into the ratio:\[ \frac{a_{n+1}}{a_n} = \frac{(2(n+1))!}{((n+1)/e)^{2(n+1)}} \times \frac{(n/e)^{2n}}{(2n)!}. \]

Simplify the Ratio

Factor in the expression for simplicity:1. Cancel \((2n)!\) in numerator and denominator.2. Separate factorials: \((2(n+1))! = (2n+2)! = (2n+2)(2n+1)(2n)!\).3. Simplify power terms: relabel and simplify the ratio of powers.Simplified form:\[ \frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1) e^2 n^{2n}}{(n+1)^{2(n+1)}}. \]

Take the Limit

Compute the limit:\[ \lim_{{n \to \infty}} \frac{(2n+2)(2n+1) e^2 n^{2n}}{(n+1)^{2(n+1)}}. \]Use the fact that \((n+1)^{2(n+1)} = (n+1)^{2n} (n+1)^2\) and \(n^{2n} = n^{2n}\) to simplify:This leads to canceling terms and examining leading coefficients.

Calculate the Resultant Limit Value

On simplifying, the limit approaches\[ \lim_{{n \to \infty}} \left[ \frac{(2 + \frac{2}{n})(2 + \frac{1}{n})}{(1 + \frac{1}{n})^2} \right] = 4.\]As \(n\) goes to infinity, this expression converges to 4.

Apply Conclusion from Ratio Test

According to the Ratio Test, if the limit \(L > 1\), the series diverges. With a final limit \(L = 4\), which is greater than 1, the original series \(\sum_{n=1}^{\infty} \frac{(2n)!}{(n / e)^{2n}} \) diverges.

Key concepts

Convergence of Series

In mathematics, the concept of convergence of series is a fundamental aspect when exploring infinite series. A series is a sum of an infinite sequence of terms, like \(\sum_{n=1}^{\infty} a_{n}\). The main question with any such sum is: does it add up to a finite number, or does it grow infinitely large?
Convergence means that as you keep adding more and more terms to the sum, the total approaches some finite value. If it does, the series is said to "converge." If not, it "diverges."

• A convergent series has a well-defined sum, even though it consists of infinitely many terms.
• A divergent series keeps increasing or decreasing without approaching a finite limit.

In practice, different tests help determine convergence. For example, the Ratio Test is one such powerful method. If using the test results in a value greater than 1, as in our example, it signals that the series diverges.

Factorial terms

Factorials, symbolized by an exclamation mark (e.g., \(n!\)), are products of all positive integers up to a specified number. For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1\).
The presence of factorial terms in a series can complicate calculations due to their rapid growth rates. \(2n!\) represents the factorial of the number 2n, meaning you multiply all whole numbers from 1 up to 2n.
This makes factorial terms especially large, and thus they heavily influence the behavior of a series.

• Any series containing factorials must be carefully examined as factorials can lead to quicker growth and often cause divergence.
• In our example, the factorial term \( (2n)! \) is central to determining whether the series converges or diverges.

The electrifying growth of factorials often makes them the deciding factor in the convergence of a series when combined with other elements like exponential terms.

Limit comparison

The Limit Comparison Test is a technique used in calculus to determine the convergence or divergence of infinite series by comparing it to a second, simpler series.
The method involves looking at the limit of the ratio of two series' terms as the sequence number goes to infinity. If this limit is a positive finite number, both series will either converge or diverge together.

• This method is especially useful when direct evaluation of a series is complex.
• It's often paired with the Ratio Test for series involving factorials or exponential expressions.

In the context of the exercise at hand, the Ratio Test simplifies the complexities by reducing it to the limit of a ratio. If you tried to compare it with a simpler series through limit comparison, you’d find it harder to handle due to the factorial and exponential terms interacting intricately.
When using the Ratio Test, identifying simplified forms that help find this limit more easily is crucial, like dividing out common terms or computing factored expressions. This can directly tell us about the series' behavior, just as we found L = 4 in this problem, verifying divergence.